Question

If the object begins to move the speed $V_0$, what will be the speed of its image formed by a spherical mirror with focal length f?

Answer 1

According to mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ (i)
For a given mirror, focal length $f=constant$. Therefore, on differentiating, Eq. (i) gives
$-\frac{1}{v_2}dv-\frac{1}{u^2}du0\Rightarrow dv=-{\left(\frac{v}{u}\right)}^{2}du$ (ii)
Multiplying Eq. (i) throughout by u, we get
$\frac{u}{v}+1=\frac{u}{f}\Rightarrow \frac{u}{v}=\frac{u}{f}-1=\frac{u-f}{f}$
$\therefore \frac{v}{u}=\frac{f}{u-f}$ (iii)
Differential length of the object on the axis, $du=b$ (given).
Therefore, Eq. (ii) gives $dv=-{\left(\frac{f}{u-f}\right)}^{2}b$.
The negative sign shows that the image is longitudinally inverted.
$\therefore$ Size of image $|dv|=b{\left(\frac{f}{u-f}\right)}^2$.
From Eqs. (ii) and (iii), $dv=-{\left(\frac{f}{u-f}\right)}^{2}du$
Dividing both the sides by differential time dt, we get
$\frac{dv}{dt}=-{\left(\frac{f}{u-f}\right)}^2\frac{du}{dt}$
Given $\frac{du}{dt}=v_0$ and speed of image $\frac{dv}{dt}=v_1$ (say)
$\Rightarrow v_i=-{\left(\frac{f}{u-f}\right)}^2{v}_0$